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Show that : $$\lim_{m\to \infty}\left[ -\frac{1}{2m}+\ln \left( \frac{\text{e}}{m} \right)+\sum\limits_{n=2}^m \left( \frac{1}{n}-\frac{\zeta \left( 1-n \right)}{m^n} \right) \right]=\gamma $$ How to evaluate that sum?

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I think the running index in the sum must be "n"...wait, or perhaps only the upper limit? This is confusing. Check the expression. – DonAntonio Jan 30 '13 at 12:17
@DonAntonio Yes, my typo – Ryan Jan 30 '13 at 12:23
And the limit? It still makes no sense... – vonbrand Jan 30 '13 at 15:52
Please don't write \underset{n\to \infty }{\mathop{\lim }}\,. I changed that to \lim_{n\to\infty}. That is standard usage. I also changed {{m}^{n}} to m^n and some other things like that. – Michael Hardy Jan 30 '13 at 17:17
I think m is the index that should go to infinity, isn't it? – Adar Hefer Jan 30 '13 at 17:21
up vote 5 down vote accepted

This is an expression of the asymptotic series for $1+\frac12+\dots+\frac1m$, $$ 1+\frac12+\dots+\frac1m= \ln{m}+\gamma+\frac{1}{2m}-\sum_{k=1}^r \frac{B_{2k}}{2k m^{2k}}+{\cal R}(m, r), $$ where $-B_{2k}/(2k)$ has been replaced by the equal value $\zeta(1-2k)$. (Since $\zeta(-s)$ is zero for $s$ positive and even, these terms can be omitted.) The term inside the limit is $$\gamma+{\cal R}(m, \lfloor m/2 \rfloor).$$

From [1], and using $1+\dots+\frac1m=\psi(m)+\gamma+\frac1m$, $$ |{\cal R}(m,r)|\le \frac{|B_{2r+2}|}{(2r+2) m^{2r+2}} \qquad (1) $$ for $m>0$ and $r\ge 0$. From [2], $$ |B_{2n}|\sim 4\sqrt{\pi n} (\frac{n}{\pi e})^{2n}, \qquad n\to\infty.\qquad (2) $$ Combining (1) and (2) gives $$ |{\cal R}(m,\lfloor m/2 \rfloor)|\le (1+o(1)) 2\sqrt{\frac{2\pi}{m}} e^{2\lfloor m/2\rfloor+2-m} (2\pi e)^{-2\lfloor m/2 \rfloor-2}.\qquad (3) $$ Since the right-hand side of (3) goes to $0$ as $m\to\infty$, this proves that the limit is $\gamma$, as desired.

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To show that $$ \lim_{m\to\infty}\left[-\frac1{2m}+\log\left(\frac em\right)+\sum_{n=2}^m \left(\frac1n-\frac{\zeta(1-n)}{m^n}\right)\right]=\gamma\tag{1} $$ we pretty simply have $$ \lim_{m\to\infty}\left[-\frac1{2m}\right]=0\tag{2} $$ and $$ \begin{align} \lim_{m\to\infty}\left[\log\left(\frac em\right)+\sum_{n=2}^m\frac1n\right] &=\lim_{m\to\infty}\left[\sum_{n=1}^m\frac1n-\log(m)\right]\\[12pt] &=\gamma\tag{3} \end{align} $$ So all that remains is to show that $$ \lim_{m\to\infty}\left[\sum_{n=2}^m\frac{\zeta(1-n)}{m^n}\right]=0\tag{4} $$ Define $$ a_m(n)=\left\{\begin{array}{} 4\sqrt{\pi/n}\left(\frac{n}{\pi me}\right)^{2n}&\text{when }n\le\lfloor m/2\rfloor\\ 4\sqrt{\pi/n}\left(\frac1{2\pi e}\right)^{2n}&\text{when }n\gt\lfloor m/2\rfloor \end{array}\right.\tag{5} $$ Note that $$ \sum_{n=1}^\infty a_0(n)\le\frac{4\sqrt\pi}{4\pi^2e^2-1}\tag{6} $$ and $a_m(n)\to0$ monotonically as $m\to0$. Thus, by Dominated Convergence, $$ \lim_{m\to\infty}\sum_{n=1}^\infty a_m(n)=0\tag{7} $$ Since $$ \begin{align} \left|\,\sum_{n=2}^m\frac{\zeta(1-n)}{m^n}\,\right| &=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}\frac{\zeta(1-2n)}{m^{2n}}\,\right|\\ &=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}-\frac{B_{2n}}{2n\,m^{2n}}\,\right|\\ &=\left|\,\sum_{n=1}^{\lfloor m/2\rfloor}(-1)^n\zeta(2n)\frac{(2n)!}{n(2\pi m)^{2n}}\,\right|\\ &\le\sum_{n=1}^{\lfloor m/2\rfloor}2\frac{\sqrt{4\pi n}(2n)^{2n}e^{-2n}}{n(2\pi m)^{2n}}\\ &=\sum_{n=1}^{\lfloor m/2\rfloor}4\sqrt{\pi/n}\left(\frac{n}{\pi me}\right)^{2n}\tag{8} \end{align} $$ we see that $(7)$ verifies $(4)$ and we are done.

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